√3sin2b+cos2b=1∴2[sin2b*(√3/2)+cos(2b)*(1/2)]=1∴2[sin2bcos(π/6)+cos2b*sin(π/6)=1∴ 2sin(2b+π/6)=1∴ sin(2b+π/6)=1/2∴ 2b+π/6=π/6或2b+π/6=5π/6∴ b=0(舍)或b=π/3利用余弦定理b²=a²+c...